Có vẻ như là đề hơi sai á bạn. Bạn xem lại đề nha.

Cho biểu thức ban đầu là A
Đặt 3 = a ; \(\sqrt{9+\dfrac{125}{27}}\)= b
⇔A = \(\sqrt[3]{a+b} . \sqrt[3]{b-a}\) 
⇔A=  \(\sqrt[3]{(a+b)(b-a)}\)
⇔A= \(\sqrt[3]{b^2-a^2}\)
⇔A= \(\sqrt[3]{9+\dfrac{125}{27}-9}\)
⇔A= \(\sqrt[3]{\dfrac{125}{27}}\)
⇔A = \(\dfrac{5}{3}\) ( ĐPCM)
 

\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

=\(\sqrt[3]{-\left(3+\sqrt{9+\dfrac{125}{27}}\right)\left(3-\sqrt{9+\dfrac{125}{27}}\right)}\)

=\(\sqrt[3]{-\left[9-\left(9+\dfrac{125}{27}\right)\right]}\)

=\(\sqrt[3]{\dfrac{125}{27}}\)

=5/3

a: \(A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot A\cdot\sqrt[3]{4-5}\)

\(\Leftrightarrow A^3=4-3A\)

=>A=1

c: \(C=1+\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(=1+3=4\)

Đề bài sai, casio cho kết quả ko phải một số nguyên, đề bài đúng phải là \(\frac{125}{27}\)

\(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Rightarrow x^3=6-3\sqrt[3]{\frac{125}{27}}\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\)

\(\Rightarrow x^3=6-5x\)

\(\Leftrightarrow x^3+5x-6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\Rightarrow x=1\)

Vậy \(x\in Z\)

a) \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

=\(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)

=\(\sqrt[3]{\left(1+\sqrt{5}\right)^3}+\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)

=\(1+\sqrt{5}+1-\sqrt{5}=2\)

b) \(\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}\)

=\(\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)

=\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)

c) xem lại đề

Bạn Thái làm sai rồi

a)do ban đầu cậu nhân 2 cho hai vế nhưng bạn chưa chia lại.mik bổ sung ý tiếp cho bạn là

2A=2=>A=1.

mik lam tiep cau b la

B=\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\)

=4-3

=1.

còn câu c mik pó tay :))

\(A=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow A=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow A^3=6+3A.\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}.\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)

\(\Leftrightarrow A^3=6+3A.\left(-\dfrac{5}{3}\right)\)

\(\Leftrightarrow A^3+5A-6=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+6\right)=0\)

\(\Leftrightarrow A=1\)

9.

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)

\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=-\sqrt{5}\)

10.

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)

\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)

\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)

\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\\ =2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}=\sqrt{5}\)

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\\ =5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}+6\sqrt{3}\\ =\dfrac{15\sqrt{3}-4\sqrt{3}+6\sqrt{6}+18\sqrt{3}}{3}\\ =\dfrac{29\sqrt{3}+6\sqrt{6}}{3}\)